Question

A spinner with 10 equal sectors numbered 1 through 10 is spun. What is the probability of the spinner randomly landing on: An even number: A prime number:A number greater than 6:2 or 5: A multiple of 3:

Answer 1

The Solution.

The set of numbers under consideration is

$\mleft\lbrace1,2,3,4,5,6,7,8,9,10\mright\rbrace=10$

Even numbers = {2,4,6,8,10} = 5

$\text{Probability(even number) =}\frac{5}{10}=\frac{1}{2}\text{ or 0.5 or 50\%}$

Prime numbers = {2,3,5,7} = 4

$\text{Probability(prime number) = }\frac{4}{10}=\frac{2}{5}\text{ or 0.4 or 40\%}$

Numbers greater than 6:

{7,8,9,10} = 4

$\text{Probability(greater than 6) = }\frac{4}{10}=\frac{2}{5}\text{ or 0.4 or 40\%}$

The probability of 2 or 5 is

$\begin{gathered} \text{Probability}(2\text{ or 5) =prob(2) + prob(5)} \\ \text{ = }\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5}\text{ or 0.2 or 20\%} \end{gathered}$

The multiple of 3:

Multiple of 3 = {3,6,9} = 3

$\text{Probability(multiple of 3) = }\frac{3}{10}\text{ or 0.333 or 33.3\%}$